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In a titration experiment 43.18 mL of 0.283 M KOH is reacted with H2SO4. The endpoint is reached when 44.62 mL of the acid is mixed with the base. What is the molar concentration of the acid?

Sagot :

The balanced formula of the reaction described is:

[tex]2KOH_{(aq)}+H_2SO_{4(aq)}→K_2SO_{4(aq)}+2H_2O_{(l)}[/tex]

To find the molar concentration (Molarity) of the solution we will follow the following steps:

1. We find the moles of KOH present in the basic solution using the molarity equation that tells us:

[tex]Molarity=\frac{MolesSolute}{Lsolution}[/tex][tex]\begin{gathered} MolesSolute=Molarity\times Lsolution \\ MolesSolute=0.283M\times0.04318L \\ MolesSolute=0.012molKOH \end{gathered}[/tex]

2. From the stoichiometry of the reaction we find the moles of H2SO4 needed to neutralize the moles of KOH. The ratio H2SO4 to KOH is 1/2.

[tex]\begin{gathered} molH_2SO_4=givenmolKOH\times\frac{1molH_2SO_4}{2molKOH} \\ molH_2SO_4=0.012molKOH\times\frac{1molH_{2}SO_{4}}{2molKOH}=0.006molH_2SO_4 \end{gathered}[/tex]

3. We find the molarity of the solution using the molarity formulation from point 1.

[tex]\begin{gathered} Molarity=\frac{MolesSolute}{Lsolution} \\ Molarity=\frac{0.006molH_2SO_4}{0.04462L}=0.137M \end{gathered}[/tex]

Answer: The molarity or molar concentration of the acid solution is 0.137M

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